# Find angle based on bisector, altitude, median info

In △ABC, CM is the median, CD is the angle bisector of ∠ACB, CH is the altitude. CM, CD, and CH divide ∠ACB by four equal angles. Find angles of △ABC.

I started with finding angle relationships but was not able to use them to proceed. Maybe this is not the right approach. Can you advise? Thanks.

Assume that $$AC. Let $$O$$ be the centre of the circumcircle of $$ABC$$. Prove that (in any triangle) angles $$\sphericalangle{BCO}=\sphericalangle{ACH}$$. Since the angle bisector already divides the angle in halves, we have to have that angle $$\sphericalangle{ACH}$$ is a quater of angle $$C$$ as well as $$\sphericalangle{BCM}$$. But since $$\sphericalangle{BCO}=\sphericalangle{ACH}$$ we must have $$\sphericalangle{BCO}=\sphericalangle{BCM}$$ i.e. points $$C$$, $$M$$, $$O$$ are collinear. Prove that it implies that $$\sphericalangle{C}=90^{\circ}$$ and usethe fact that then $$\sphericalangle{ACH}=\frac{90^{\circ}}{4}$$ to compute the rest.
Let $$\measuredangle ACH=\alpha$$.
Thus, $$\measuredangle A=90^{\circ}-\alpha$$ and $$\measuredangle B=90^{\circ}-3\alpha.$$
Thus, by the law of sines we obtain: $$\frac{\sin3\alpha}{\cos\alpha}=\frac{AM}{MC}=\frac{BM}{MC}=\frac{\sin\alpha}{\cos3\alpha},$$ which gives $$\sin6\alpha=\sin2\alpha$$ or $$6\alpha+2\alpha=180^{\circ}.$$ Can you end it now?